# I need some help on my math homework and I have no clue on how to go about it.?

Two Airplanes leave the same airport in the opposite directions. At 2:00P.M the angle of elevation from the aiport to the first plane is 26 degrees and to the 2nd plane is 17 degrees. The first plane has an elevation of 7km. The 2nd plane has an elevation of 8.2km find the line of sight distance between the 2 airplanes.

pls help :c

Relevance
• Horizontal distance between two planes = hor. dist. of 1st plane + hor. dist. of 2nd plane

hor. dist. = 7 x cos 26° + 8.2 x cos 17° = 14.1332 Km

Vertical distance difference between two planes = 8.2 - 7 = 1.2 Km

by Pythagorean Theorem

sight distance = ( 14.332 ² + 1.2 ² ) ^ (1/2) = 14.3821 Km

• Alan
Lv 7
5 months agoReport

Elevation is 7 km and 8.2 km
7 is opposite the 26 degree angle

• Think of the y axis as elevation, and the positive x axis as the (ground) direction of the first plane. Then its coordinates at 2 p.m. are (7 cot(26), 7) and the coordinates of the 2nd plane are (-8.2 cot(17), 8.2). So the distance between the two airplanes is

sqrt{[7 cot(26) + 8.2 cot(17)]^2 + (8.2 - 7)^2}

= 41.19 km

• Let the aircraft be a and b

Angle between the aircraft = 180-26-17 = 137°

Slant height of first aircraft = Sa

Slant height of second aircraft = Sb

Sa = 7/sin26 = 15.9682 km

Sb = 8.2/sin17 = 28.0465 km

Let x be the required distance

Cosine Rule

x^2 =

15.9682^2 + 28.0465^2 - 2*15.9682*28.0465cos137

x = 41.19 km

• The law of cosines can be used to calculate the line of sight (LOS) distance, c, between the two planes. You know the angle opposite that distance Θ = 180 - 26 - 17 = 137­°. What's missing is the LOS distance to each plane. But you are given all that's needed to find those distances.

For the first plane, you know the altitude and elevation angle, 7 km and 17°. The LOS distance, a, can be found from the trigonometry of the situation:

a = 7/sin17°

For the second plane, the configuration is altitude 8.2 km and elevation angle 28­°. The LOS distance, b, to the second plane from the airport is:

b = 8.2/sin28°

That's all you need to apply the law of cosines to the problem:

c² = a² + b² - 2ab cosΘ

= (7/sin17)² + (8.2/sin28)² - 2(7/sin17)(8.2/sin28) cos 137

= 573.225 + 305.077 + 611.680

= 1489.983

c = 38.6 km

• Alan
Lv 7
5 months agoReport

Math Typoes
sin(26) changed to sin(28) why.
You used 26 and 17 to find the 137
then changed 26 to 28 (for no good reason.)

• Sketch the situation.....

then

if x is the horizontal distance from the airport for plane 1

tan(26) = 7/x

==> x = 7/tan(26)

if y is the distance from the airport for plane 2

tan(17) = 8.2/y

==> y = 8.2/tan(17)

line of sight (call it w) between the panes is thus

(x+y)^2 + (8.2 - 7)^2 = w^2

solve for w.....

• Alan
Lv 7
5 months agoReport

No, what you just said is exactly
the same as the distance formula. The distance formula
is totally based on the Pythagorean theorem
(They aren't different.)
distance = sqrt( (y - (-x))^2 + (8.2-7)^2 )
distance = sqrt( (y+x)^2 + (8.2 -7)^2 )
same thing as what you just said, not different.